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Question

The tangents to the parabola y2=4ax at P(t1) and Q(t2) intersect at R. Then the area of â–³PQR

A
a22(t1t2)2
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B
a22(|t1t2|)
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C
a22(|t1t2|)3
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D
a2(t1t2)2
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Solution

The correct option is B a22(|t1t2|)
y2=4ax........(i)
Tangents are drawn at P(t1)andQ(t2)
Intersects at R.
Points of intersection of tangents at P(t2)andQ(t2)
=>R:(at1t2,a(t1+t2))
Now tangent are drawn from R to equation(i)
So, area of triangle formed by then tangent and chord of contact.
When tangents are drawn from (x1,y1)=(y214ax1)32.12a
Here, x1=at1t2,y1=a(t1+t2)
=>AreaofRPQ=12a[(a(t1+t2]24a(at1t2)]32
=12a(a2)32[(t1+t2)24t1t2]32
=a22(t1t2)2]32(a2+b22ab=(a2+b2)24ab)
=a22[|t1t2|]3

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